2.2等差數(shù)列的前n項和第1課時等差數(shù)列的前n項和課后訓(xùn)練鞏固提升1.設(shè)Sn為等差數(shù)列{an}的前n項和,若a1=1,公差d=2,Sk+2-Sk=24,則k等于().A.8 B.7 C.6 D.5解析:∵Sk+2-Sk=ak+1+ak+2=a1+kd+a1+(k+1)d=2a1+(2k+1)d=2×1+(2k+1)×2=4k+4=24,∴k=5.答案:D2.已知等差數(shù)列{an}的前n項和為Sn,且S3=6,a1=4,則公差d等于().A.1 B.53 C.-2 D.解析:由題意,得6=3a1+12×3×2×d,又a1=故d=-2.答案:C3.已知等差數(shù)列{an}的公差為d,前n項和為Sn,當(dāng)首項a1和d變化時,a2+a8+a11是一個定值,則下列各數(shù)中也為定值的是().A.S7 B.S8 C.S13 D.S15解析:由已知a2+a8+a11=3a1+18d=3(a1+6d)=3a7,且為定值,則S13=13(a1+a13)2=13答案:C4.設(shè)Sn為等差數(shù)列{an}的前n項和,S3=4a3,a7=-2,則a9=().A.-6 B.-4 C.-2 D.2解析:S故a9=a1+8d=-6.答案:A5.已知等差數(shù)列{an}共有10項,其中奇數(shù)項之和為15,偶數(shù)項之和為30,則其公差是().A.5 B.4 C.3 D.2解析:設(shè)公差為d,則a∴5d=15,∴d=3.答案:C6.設(shè)Sn是等差數(shù)列{an}的前n項和,若a7a4=2,則S13解析:S13答案:267.設(shè)數(shù)列{an}的前n項和為Sn,點n,Snn(n∈N+)均在函數(shù)y=3x-2的圖象上,則數(shù)列{an}的通項公式為.

解析:依題意得,Snn=3n-2,即Sn=3n2-2當(dāng)n≥2時,an=Sn-Sn-1=(3n2-2n)-[3(n-1)2-2(n-1)]=6n-5.因為a1=S1=1,滿足an=6n-5,所以an=6n-5(n∈N+).答案:an=6n-5(n∈N+)8.已知數(shù)列{an}滿足a1+2a2+…+nan=n(n+1)(n+2),則an=.

解析:由a1+2a2+…+nan=n(n+1)(n+2),①得a1+2a2+…+(n-1)an-1=(n-1)n(n+1),②由①-②,得nan=n(n+1)(n+2)-(n-1)n(n+1)=n(n+1)[(n+2)-(n-1)]=3n(n+1),an=3(n+1)(n≥2).當(dāng)n=1時,a1=1×2×3=6,也適合上式,∴an=3(n+1),n∈N+.答案:3(n+1)9.已知數(shù)列{an}是等差數(shù)列,a2=5,a5=14.(1)求數(shù)列{an}的通項公式;(2)設(shè)數(shù)列{an}的前n項和Sn=155,求n的值.解:(1)設(shè)等差數(shù)列{an}的公差為d,首項為a1,則a∴數(shù)列{an}的通項公式為an=3n-1.(2)數(shù)列{an}的前n項和Sn=32n2+12由32n2+12n=155,可得n=1010.已知公差大于零的等差數(shù)列{an}的前n項和為Sn,且滿足:a3a4=117,a2+a5=22.(1)求數(shù)列{an}的通項公式an;(2)若數(shù)列{bn}是等差數(shù)列,且bn=Snn+c,解:(1)設(shè)等差數(shù)列{an}的首項為a1,公差為d,且d>0.∵a3+a4=a2+a5=22,又a3a4=117,公差d>0,∴a3<a4,∴a3=9,a4=13.∴a∴an=4n-3.(2)由(1)知,Sn=n×1+n(n-1)2×4∴bn=Sn∴b1=11+